Check sibling questions

Example 10 - Find term independent of x in (3/2 x^2 - 1/3x)^6

Example  10 - Chapter 8 Class 11 Binomial Theorem - Part 2
Example  10 - Chapter 8 Class 11 Binomial Theorem - Part 3
Example  10 - Chapter 8 Class 11 Binomial Theorem - Part 4

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Transcript

Example 10 Find the term independent of x in the expansion of (3/2 π‘₯^2 " βˆ’ " 1/3π‘₯)^6,x > 0. Calculating general term We know that general term of expansion (a + b)n is Tr + 1 = nCr (a)n–r.(b)n For general term of expansion (3/2 π‘₯^2 " βˆ’ " 1/3π‘₯)^6 Putting n = 6 , a = 3/2 π‘₯^2 , b = "βˆ’" 1/3π‘₯ Tr + 1 = 6Cr (πŸ‘/𝟐 𝒙^𝟐 )^(πŸ” βˆ’ 𝒓) ((βˆ’πŸ)/πŸ‘π’™)^𝒓 = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) (π‘₯^2 ")" ^((6 βˆ’ π‘Ÿ)) ((βˆ’1)/3 "Γ—" 1/π‘₯)^π‘Ÿ = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) (π‘₯")" ^(2(6 βˆ’ π‘Ÿ)) ((βˆ’1)/3)^π‘Ÿ (1/π‘₯)^π‘Ÿ = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) (π‘₯")" ^(12 βˆ’ 2π‘Ÿ) ((βˆ’1)/3)^π‘Ÿ (π‘₯)^(βˆ’π‘Ÿ) = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) ((βˆ’1)/3)^π‘Ÿ (π‘₯")" ^(12 βˆ’ 2π‘Ÿ) (π‘₯)^(βˆ’π‘Ÿ) = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) ((βˆ’1)/3)^π‘Ÿ (π‘₯")" ^(12 βˆ’ 2π‘Ÿ βˆ’ π‘Ÿ) = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) ((βˆ’1)/3)^π‘Ÿ (π‘₯")" ^(12 βˆ’3π‘Ÿ) We need to find the term independent of x So, power of x is 0 π‘₯^(12 βˆ’ 3π‘Ÿ) = x0 Comparing powers 12 – 3r = 0 12 = 3r 12/3 = r 4 = r r = 4 Putting r = 4 in (1) T4+1 = 6C4 (3/2)^(6 βˆ’ 4) ((βˆ’1)/3)^4 (π‘₯")" ^(12 βˆ’3(4)) T5 = 6C4 (3/2)^2 (1/3^4 ) (π‘₯")" ^(12 βˆ’12) = 6C4 (3^2/2^2 )(1/3^4 ) (π‘₯")" ^0 = 6C4 (1/2^2 )(3^2/3^4 ) (1) = 6C4 (1/2^2 )(1/3^2 ) = 6!/4!(6 βˆ’ 4)! (1/4)(1/9) = 6!/4!(2)! (1/4)(1/9) = (6(5)(4)!)/4!(2)! (1/4)(1/9) = (5 )/12 Hence, the term which is independent of x is 5th term = T5 = πŸ“/𝟏𝟐

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.